In order to find the value I took $F =\mathbb{Q}(\sqrt{2})$.
So,$F(\sqrt{6}) = \{a + b\sqrt{6} | a,b \in \mathbb{Q}(\sqrt{2})\}$
So expanding the elements $a$ and $b$ as elements of $\mathbb{Q}(\sqrt{2})$, $= (a_{1} + b_{1}\sqrt{2}) +(a_{2}+b_{2}\sqrt{2})\sqrt{6}$
Solving a bit more gives me
$a_{1} + b_{1}\sqrt{2} + \sqrt{3}(a_{2}\sqrt{2} + 2b_{2})$.
Now are the basis elements $1$ and $\sqrt{2}$ ?
Now how do I choose the basis elements in order to find the dimension of the vector space $\mathbb{Q}(\sqrt{2})(\sqrt{6})$ over $\mathbb{Q}(\sqrt{3})$ ?
Any help is great!